Question

1. in R3, let U span (1,0,0),(0,1/root2, 1/root2). then U is an element of U such that ||u-(2,4,6)|| is as small as possible.
2. for a given function f:R to R defined as f(x)=2x-1, the image of S={x is an element of R: x^2-4>/=0} is?
3. suppose T: R^2 to M22 is a linear defined by T(U,V)=[(U,U), (V,2U)]. Then ker(T) is?
4. suppose T:R^6 to R^4 is a linear map such that null T=U, where U is 2-dimensional subspace of R^6. Then dim range T is?

Answer 1

1.

$A=\begin{pmatrix} 1 & 0 \\ 0 & 1/\sqrt 2\\ 0&1/\sqrt 2 \end{pmatrix},b=\begin{pmatrix} 2 \\ 4\\ 6 \end{pmatrix}$

Then u is the orthogonal projection of b onto the subspace spanned by the column of A and it is given by the formula:

$u=A(A^TA)^{-1}A^Tb$

then:

$A^Tb=\begin{pmatrix} 1 & 0&0 \\ 0 & 1/\sqrt 2& 1/\sqrt 2\\ \end{pmatrix}\begin{pmatrix} 2 \\ 4\\ 6 \end{pmatrix}=\begin{pmatrix} 2 \\ 10/\sqrt 2 \end{pmatrix}$

$A^TA=\begin{pmatrix} 1 & 0&0 \\ 0 & 1/\sqrt 2& 1/\sqrt 2\\ \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 1/\sqrt 2\\ 0&1/\sqrt 2 \end{pmatrix}=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

$(A^TA)^{-1}=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

$(A^TA)^{-1}A^Tb=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 2 \\ 10/\sqrt 2 \end{pmatrix}=\begin{pmatrix} 2 \\ 10/\sqrt 2 \end{pmatrix}$

$u=\begin{pmatrix} 1 & 0 \\ 0 & 1/\sqrt 2\\ 0&1/\sqrt 2 \end{pmatrix}\begin{pmatrix} 2 \\ 10/\sqrt 2 \end{pmatrix}=\begin{pmatrix} 2 \\ 5\\ 5 \end{pmatrix}$

2.

$x^2-4\ge 0 \implies |x|\ge 2$

then image of S:

$f(x)\le 2(-2)-1=-5$ and

$f(x)\ge 2\cdot 2-1=3$

3.

for ker(T):

$U+V=0$

$U+2U=0$

so,

$ker(T)=(0,0)$

4.

by the rank-nullity theorem:

$6=dim(Ker(T))+dim(Im(T))$

So, dim range T= 6-dim(Ker(T))=6-2=4